Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 1e5 + 5;
int n, m, q;
vector<int> adj[N], c[N];
int anc[N][20];
int h[N];
vector<int> store[N][20];
vector<int> Merge(vector<int> a, vector<int> b) {
  vector<int> ans(0);
  for (int i = 0, j = 0; (i < ((int)(a).size()) || j < ((int)(b).size())) &&
                         ((int)(ans).size()) <= 10;) {
    if (i == ((int)(a).size()))
      ans.push_back(b[j++]);
    else if (j == ((int)(b).size()))
      ans.push_back(a[i++]);
    else if (a[i] == b[j]) {
      ans.push_back(a[i]);
      i++;
      j++;
    } else if (a[i] < b[j])
      ans.push_back(a[i++]);
    else
      ans.push_back(b[j++]);
  }
  return ans;
}
vector<int> Find(int u, int cha) {
  int k = log2(h[u]);
  vector<int> ans;
  for (int i = k; i >= 0; --i) {
    if (h[u] - (1 << i) >= h[cha]) {
      ans = Merge(ans, store[u][i]);
      u = anc[u][i];
    }
  }
  ans = Merge(ans, c[cha]);
  return ans;
}
void dfs(int u, int pre) {
  for (int v : adj[u])
    if (v != pre) {
      h[v] = h[u] + 1;
      anc[v][0] = u;
      store[v][0] = Merge(store[v][0], c[u]);
      dfs(v, u);
    }
}
void initLCA() {
  for (int j = 1; j <= log2(n); ++j) {
    for (int i = 1; i <= n; ++i) {
      anc[i][j] = anc[anc[i][j - 1]][j - 1];
      store[i][j] = Merge(store[i][j - 1], store[anc[i][j - 1]][j - 1]);
    }
  }
}
int LCA(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int k = (int)log2(h[u]);
  for (int i = k; i >= 0; --i) {
    if (h[u] - (1 << i) >= h[v]) {
      u = anc[u][i];
    }
  }
  if (u == v) return u;
  for (int i = k; i >= 0; --i) {
    if (anc[u][i] && anc[u][i] != anc[v][i]) {
      u = anc[u][i];
      v = anc[v][i];
    }
  }
  return anc[u][0];
}
void gogo() {
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    int x;
    cin >> x;
    store[x][0].push_back(i);
    c[x].push_back(i);
  }
  dfs(1, 1);
  initLCA();
  while (q--) {
    int u, v, lim;
    cin >> u >> v >> lim;
    int cha = LCA(u, v), near_cha = v;
    vector<int> ans = Merge(Find(u, cha), Find(v, cha));
    cout << min(((int)(ans).size()), lim) << ' ';
    for (int i = 0; i < min(((int)(ans).size()), lim); ++i)
      cout << ans[i] << ' ';
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
}